proof of square root of square root binomial

We square the expression on the right-hand-side and expand
using the binomial formula:

(a+a2-b2±a-a2-b2)2

=(a+a2-b2)2

+(a-a2-b2)2±2⁢a+a2-b2⁢a-a2-b2

Since the squaring operation undoes the square roots, we
obtain the following:

(a+a2-b2)2+(a-a2-b2)2=a+a2-b2+a-a2-b2=a

Since the product of square roots equals the square root
of the product, we have the following:

a+a2-b2⁢a-a2-b2

=a+a2-b2⋅a-a2-b2

=a2-(a2-b)24

=a2-(a2-b)4

=b4=b2

Combining what we have calculated above, we obtain

(a+a2-b2±a-a2-b2)2=a±b.

Because the square of the asserted value of the square root equals the radicand (a±b) of the square root, and the asserted value of the square root is clearly non-negative, we have justified the validity of the formulas